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Journal of Integer Sequences, Vol. 12 (2009),
Article 09.1.5
23 11
The Pfaffian Transform
Tracale Austin
3824 Teton Pass
Ellenwood, GA 30294
USA
austintra@gmail.com
Hans Bantilan
Department of Physics
Princeton University
Princeton, NJ 08544
USA
bantilan@princeton.edu
Eric S. Egge
Department of Mathematics
Carleton College
Northfield, MN 55057
USA
eggee@member.ams.org
Isao Jonas
Challenge Online Games, Inc.
816 Congress Ave, Suite 1470
Austin, TX 78701
USA
isao.jonas@gmail.com
Paul Kory
Department of Mathematics
Indiana University
831 East 3rd Street
Bloomington, IN 47405
USA
pkory@indiana.edu
1
Abstract
We introduce a function on sequences, which we call the Pfaffian transform, using
the Pfaffian of a skew-symmetric matrix. We establish several basic properties of the
Pfaffian transform, and we use the transfer matrix method to show that the set of
sequences with rational generating functions is closed under the Pfaffian transform.
We conclude by computing the Pfaffian transform of a variety of sequences, including
geometric sequences, the sequence of Fibonacci numbers, the sequence of Pell numbers, the sequence of Jacobsthal numbers, and the sequence of Tribonacci numbers.
Throughout we describe a generalization of our results to Pfaffians of skew-symmetric
matrices whose entries satisfy a Pascal-like relation.
1
Introduction
The Hankel transform of a sequence {an }∞
n=0
nth term is the Hankel determinant

a0 a1
 a1 a2


hn = det  a2 · · ·
 ..
..
.
.
an · · ·
∞
is the sequence H({an }∞
n=0 ) = {hn }n=0 whose

a2 · · ·
an
· · · · · · an+1 

· · · · · · an+2 
.
.. 
..
..
. 
.
.
· · · · · · a2n+2
The Hankel transform was first introduced by Layman [11], who showed that the Hankel
transform of a sequence is equal to the Hankel transform of both the Binomial and Invert
transforms of that sequence. Layman’s work on the Hankel transform has been extended
in [4, 5, 12]. The study of determinants of Hankel matrices predates the introduction of
the Hankel transform, so the Hankel transforms of many sequences were already known
2n
1
is the
when [11] appeared. For instance, the sequence of Catalan numbers Cn = n+1
n
∞
∞
unique sequence for which H({Cn }n=0 ) and H({Cn }n=1 ) are both the sequence consisting
entirely of 1s [13, Ex. 6.26b], and Desainte-Catherine and Viennot have shown [6] that
Y
i + j + 2m
. Generalizing
when k ≥ 2 the sequence H({Cn }∞
n=k ) has mth term
i+j
1≤i≤j≤k−1
the Catalan result in a different direction, Aigner has shown [1] that the Motzkin numbers
∞
∞
have H({Mn }∞
n=0 ) = {1}n=0 , while H({Mn }n=1 ) is the sequence of period six which begins
1, 0, −1, −1, 0, 1. Aigner provides a generalization of these results in [2]. Other results
concerning Hankel determinants and the Hankel transform can be found in [3, 7, 15, 16],
though this list is not exhaustive.
The determinant of an n × n matrix, which plays a central role in the Hankel transform,
can be defined as a sum over the set of perfect matchings in the complete bipartite graph
Kn,n . Here a perfect matching in a graph G is a set of edges in G such that every vertex is
incident with exactly one edge in the set. For n × n skew-symmetric matrices, one natural
analogue of the determinant is the Pfaffian, which is a sum over perfect matchings in the
complete graph Kn . Like determinants, Pfaffians have been connected with a variety of
2
combinatorial objects, including plane partitions (through families of nonintersecting lattice
paths) [14], the Bender-Knuth conjecture [9], and the matrix-tree theorem [10].
In this paper we introduce an analogue of the Hankel transform, called the Pfaffian
transform, which uses the Pfaffian of a skew-symmetric Toeplitz matrix in place of the
determinant of a Hankel matrix. Although the Pfaffian transform may be applied to any
sequence, we study its action on sequences with rational generating functions in particular.
We first use standard matrix operations to show that computing the Pfaffian transform of a
sequence with a rational generating function boils down to computing the Pfaffian transform
of a sequence which is eventually 0. We then use the transfer matrix method to show that the
Pfaffian transform of any such sequence has a rational generating function. This, in turn,
enables us to show that the set of sequences with rational generating functions is closed
under the Pfaffian transform. We conclude the paper by computing the Pfaffian transform
on a variety of specific sequences, some of which are given in the table below.
{an }∞
n=1
1, 2, 4, . . . , 2n−1 , . . .
1, 3, 9, . . . , 3n−1 , . . .
1, 1, 2, 3, . . . , Fn , . . .
1, 1, 3, 5, 11, 21, . . . , Jn , . . .
1, 1, 2, 4, 7, 13, . . . , Tn , . . .
Pfaffian transform of {an }∞
n=1
1, 1, 1, . . .
1, 1, 1, . . .
1, 2, 4, . . . , 2n−1 , . . .
1, 3, 9, . . . , 3n−1 , . . .
1, 2, 3, . . . , n, . . .
Here Fn is the nth Fibonacci number, Jn is the nth Jacobsthal number (these satisfy Jn =
Jn−1 + 2Jn−2 ), and Tn is the nth Tribonacci number (these satisfy Tn = Tn−1 + Tn−2 + Tn−3 ).
Our examples are, in fact, more general than the table above suggests. For instance, we
show that for any c 6= 0 the Pfaffian transform of {cn−1 }∞
n=1 is 1, 1, 1, . . ., and we show that
for any sequence {an }∞
which
satisfies
a
=
1,
a
=
c,
and an = can−1 + an−2 for n ≥ 3,
1
2
n=1
∞
where c is arbitrary, the Pfaffian transform of {an }n=1 is 1, 2, 4, . . . , 2n−1 , . . .. In our last
example we show that for certain sequences the Pfaffian transform can be given in terms of
the matchings polynomial of a path.
2
The Pfaffian Transform of a Sequence
In this section we introduce the Pfaffian transform of a sequence, which is defined using
the Pfaffian of a skew-symmetric matrix. The Pfaffian of a skew-symmetric matrix is given
in terms of a certain graph, so we begin by adopting some graph theoretic conventions.
Throughout we use graphs with vertex set [n] := {1, 2, . . . , n}, and we write edges in a graph
with their smaller vertex first. We recall that the complete graph with n vertices is the graph
in which each pair of vertices is connected by an edge; we write Kn to denote this graph.
We also recall that a perfect matching α in a graph G is a set of edges of G such that every
vertex is contained in exactly one edge in α; we write α G to indicate that α is a perfect
matching in G. With these conventions in mind, we turn our attention to the Pfaffian of a
skew-symmetric matrix.
Definition 1.
Suppose G is a graph with edges (i, j) and (k, l). We say (i, j) and (k, l)
are crossed whenever i < k < j < l or k < i < l < j. For any perfect matching α in G,
3
we write cr(α) to denote the crossing number of α, which is the number of pairs of crossed
edges in α.
The Pfaffian of an n × n skew-symmetric matrix is a sum over perfect matchings in Kn .
When n is odd no such perfect matchings exist, so we define the Pfaffian only when n is
even.
Definition 2.
Suppose S is a 2n × 2n skew-symmetric matrix. For any perfect matching
α K2n , we write wt(α) to denote the weight of α, which is defined by
Y
wt(α) = (−1)cr(α)
Si,j .
(i,j)∈α
We also write Pf(S) to denote the Pfaffian of S, which is defined by
X
Pf(S) =
wt(α).
(1)
αK2n
The Pfaffian is closely related to the determinant, and in particular one can show that if
we replace K2n with the complete bipartite graph with parts {1, 2, . . . , 2n} and {1, 2, . . . , 2n}
in (1) then we obtain det(S) instead of Pf(S). In view of this, it’s not surprising that the
Pfaffian behaves well with respect to certain products of matrices, as we show next.
Proposition 3. Suppose S is a 2n×2n skew-symmetric matrix and P is an arbitrary 2n×2n
matrix. Then P SP t is skew-symmetric and
Pf(P SP t ) = det(P ) Pf(S).
(2)
Proof. First note that (P SP t )t = P S t P t = −P SP t , so P SP t is skew-symmetric.
To prove (2), first observe that by Definition 2 we have
Pf(P SP t ) =
X
2n
2n X
Y X
(−1)cr(α)
Pi,k Sk,m Pj,m .
(3)
(i,j)∈α k=1 m=1
αK2n
Now note that for each α K2n , the associated term in (3) is a product of n entries of S
and exactly one entry from each row of P . Therefore we have
2n
2n X
Y X
(i,j)∈α k=1 m=1
Pi,k Sk,m Pj,m =
n
XY
π
r=1
Pr,π(r)
Y
Sπ(i),π(j) ,
(4)
(i,j)∈α
where the sum on the right is over all functions from [2n] to [2n]. If π is such a function
and there exists an edge (i, j) ∈ α with π(i) = π(j) then its associated term is zero, since
S is skew-symmetric. More generally, we claim that if π is not injective, then its associated
term cancels from this sum. To prove this, we construct a sign-reversing involution on such
terms.
Fix a function π from [2n] to [2n] which is not injective, but for which each edge (i, j) ∈ α
has π(i) 6= π(j). Let a denote the smallest number which is repeated in the image of π. Now
4
let i denote the smallest number in [2n] which maps to a under π or which is matched with
such a number via an edge in α, and let (i, j) denote the edge in α which contains i. Define
π ′ to be the function from [2n] to [2n] given by


π(i), if k = j;
′
π (k) = π(j), if k = i;


π(k), otherwise.
By construction the map π 7→ π ′ is an involution, and since S is skew-symmetric, the terms
associated with π and π ′ are negatives of one another, and thus cancel. This completes the
proof of the claim. This allows us to take the sum on the right side of (4) to be over S2n ,
the set of permutations of [2n].
To complete the proof of the Proposition,
note that switching two numbers r and s in
Y
cr(α)
[2n] reverses the sign of (−1)
Si,j , by reversing the sign of Si,j if r and s share an
(i,j)∈α
edge in α and by changing cr(α) by one if they don’t. Therefore
X
Pf(P SP t ) =
(−1)cr(α)
α
= det(P ) Pf(S),
X
π∈S2n
(−1)inv(π)
2n
Y
r=1
Pr,π(r)
!
Y
Si,j
(i,j)∈α
as desired.
As we show next, Proposition 3 allows us to perform row and column operations on
a skew symmetric matrix without changing its Pfaffian, as long as we perform the same
operations on the columns as on the rows.
Corollary 4. Let S be a 2n × 2n skew-symmetric matrix and let c be a constant. If T is the
matrix obtained from S by adding c times row s to row r and c times column s to column r,
where 0 ≤ r < s ≤ 2n, then Pf(T ) = Pf(S).
Proof. Consider the row and column operations in terms of matrix multiplication. The row
operation is equivalent to multiplying S by a matrix B on the left, where Brs = c, B has
ones on the main diagonal, and zeros everywhere else. Similarly, the column operation is
equivalent to multiplying S by B t on the right. Thus T = BSB t . Since B is an uppertriangular matrix, it is easy to see that det(B) = 1. Therefore by Proposition 3 we see that
Pf(T ) = Pf(S), as desired.
As Corollary 4 suggests, in practice one can compute the Pfaffian of a given skewsymmetric matrix using Gaussian elimination, much as one can compute a determinant.
Although we do not use it here, there is also a Pfaffian analogue of the standard Laplace
expansion of a determinant [8].
We are now ready to define the Pfaffian transform.
5
Definition 5.
For any sequence {an }∞
n=1 and any k ≥
symmetric matrix given by

0
a1
a2
a3
 −a1
0
a1
a2

 −a2
−a
0
a
1
1

Sk =  −a
−a
−a
0
3
2
1

 ..
..
..
..
 .
.
.
.
−a2k−1 −a2k−2 −a2k−3 −a2k−4
1, let Sk be the 2k × 2k skew
a2k−1
a2k−2 

a2k−3 

.
a2k−4 


..
. 
···
0
···
···
···
···
...
(5)
∞
The Pfaffian transform of {an }∞
n=1 is Pf({an }) = {Pf(Sk )}k=1 . In order to facilitate discussion of individual elements of the output we also abbreviate ãk = Pf(Sk ). Thus we have
Pf({an }) = {ãn }.
Note that we use the notation Pf() for both the Pfaffian of a matrix and the Pfaffian
transform of a sequence. This is not ambiguous so long as we are careful with our parentheses, but as an additional aid to clarity we generally use uppercase letters for matrices and
lowercase letters for scalars.
We conclude this section by showing that the Pfaffian transform behaves well under
scaling by a constant.
Proposition 6. For any sequence {an }∞
˜ k = ck ãk for all
n=1 and any constant c, we have ca
k ≥ 1.
′
Proof.
√ Let Sk denote the matrix obtained
√by replacing
√ t ai with cai for 1 ≤ i ≤ 2k − 1 in (5).
′
If c is any square root of c then Sk√= ( cI)Sk ( cI) , where I is the 2k × 2k identity. By
Proposition 3 we have Pf(Sk′ ) = det( cI) Pf(Sk ), and the result follows.
3
Reducing Long Sequences to Short Sequences
For the rest of this paper we study the action of the Pfaffian transform on sequences with
rational generating functions, which are exactly those sequences which (eventually) satisfy a
linear homogeneous recurrence relation with constant coefficients. In this section we describe
an algorithm that reduces such an input sequence to a sequence which is eventually zero,
without changing the image under the Pfaffian transform. We begin with some terminology.
Definition 7.
We say an n × n matrix S is banded whenever there exists m with m < n
such that Sij = 0 whenever |i − j| > m.
The diagonal matrices are the banded matrices with m = 1, while banded matrices with
m = 2 are often called tri-diagonal matrices.
Definition 8.
Fix integers k, n, N ≥ 1 and a sequence {am }∞
m=1 , and suppose there are
constants β1 , . . . , βN for which
am =
N
X
(m ≥ N + k).
βi am−i
i=1
6
Let S denote the n × n skew-symmetric matrix for which Si,j = aj−i whenever j > i. We
write red(S) to denote the n × n matrix obtained by performing the following operations on
the entries of S.
1. For each i from 1 to n, do the following. For each j from 1 to n, replace the current
N
X
entry Si,j with Si,j −
βm Si+m,j .
m=1
2. For each j from 1 to n, do the following. For each i from 1 to n, replace the current
N
X
entry Si,j with Si,j −
βm Si,j+m .
m=1
Note that step 1 in this definition amounts to subtracting, from each row of the current
matrix S, a sum of all subsequent rows, in which the terms are weighted by the coefficients in
the given recurrence relation. Similarly, step 2 amounts to subtracting, from each column of
the current matrix S, a sum of all subsequent columns, in which the terms are again weighted
by the coefficients in the given recurrence relation. As we show next, these operations simplify
the matrix considerably without changing its Pfaffian.
Proposition 9. Fix k, n, N ≥ 1 and suppose {am }∞
m=1 satisfies
am =
N
X
(m ≥ N + k).
βi am−i
i=1
(6)
Let S denote the n × n skew-symmetric matrix for which Si,j = aj−i whenever j > i. Then
the following hold.
(i) The matrix red(S) is a banded skew-symmetric matrix with at most N + k nonzero
bands.
(ii) The entries in each band of red(S) are constant, except possibly in an N × N submatrix
in the bottom-right corner.
(iii) Pf(red(S)) = Pf(S).
Proof. First observe that the operations used to obtain red(S) are symmetric row and column operations of the type described in Corollary 4, so red(S) is skew-symmetric and
Pf(red(S)) = Pf(S).
To show that red(S) is banded with nearly constant bands, first consider the entries of
red(S) above its main diagonal. For convenience, let B denote the matrix obtained from
step 1 of Definition 8. If j − i ≥ N + k then
Bi,j = Si,j −
N
X
= aj−i −
= 0,
7
βm Si+m,j
m=1
N
X
m=1
βm aj−i−m
where the last step follows
PN from (6). When we apply step 2 of Definition 8 to the ijth entry
of B we obtain Bi,j − m=1 βm Bi,j+m . But j − i ≥ N + k so j + m − i ≥ N + k; therefore
each term in this sum is 0. It follows that the entries above the main diagonal of red(S) are
banded. But red(S) is skew-symmetric, so red(S) is banded with at most N + k nonzero
bands.
Now observe that the algorithm which produces red(S) performs the same operations on
all entries in a given band except for those entries that are N rows away from the last row or
N columns away from the last column. Therefore the bands are constant, with the possible
exception of a submatrix in the bottom-right corner of size at most N × N .
We conclude this section by describing how one can generalize Proposition 9 to a wider
class of skew-symmetric matrices.
Definition 10.
Fix complex numbers ν, λ, and β1 , . . . , βN . We say a 2n × 2n skewsymmetric matrix S is a ν, λ-Pascal matrix with coefficients β1 , . . . , βN whenever the following
hold.
P
1. S2n−1,2n = 1 and for all j, 1 ≤ j ≤ 2n − 2, we have Sj,2n = N
i=1 βi Sj+i,2n . Here we set
Si,j = 0 if i > 2n.
2. For all i, j, 1 ≤ i, j ≤ 2n − 1, we have Si,j = λSi+1,j+1 + ν(Si+1,j + Si,j+1 ).
As we describe next, we can use an analogue of red to turn a ν, λ-Pascal matrix S into
a banded skew-symmetric matrix with nearly constant bands.
Definition 11.
Fix complex numbers ν, λ, and β1 , . . . , βN and suppose S is a ν, λ-Pascal
matrix with coefficients β1 , . . . , βN . We write Pred(S) to denote the matrix obtained from
S as follows.
1. Replace S with red(S).
2. For each k from N + 1 to 2n − 1, replace the current matrix S with a new matrix S,
as follows.
(a) For each i from 1 to 2n − k, do the following. For each j from 1 to 2n, replace
Si,j with Si,j − νSi+1,j .
(b) For each j from 1 to 2n − k, do the following. For each i from 1 to 2n, replace
Si,j with Si,j − νSi,j+1 .
3. For all i, j, 1 ≤ i, j ≤ 2n − 2, replace the current entry Si,j with ω
ω = λ + ν 2.
3+i+j
−2n
2
Si,j , where
If S is a ν, λ-Pascal matrix with coefficients β1 , . . . , βN , then each step in the computation
of Pred(S) can be accomplished by replacing the current matrix S with P SP t for a certain
2
upper-triangular matrix P . In the last step det(P ) = ω −(n−1) , where ω = λ + ν 2 , but in all
2
of the other steps det(P ) = 1. As a result, Pf(Pred(S)) = ω −(n−1) Pf(S) by Proposition 3.
In fact, one can also prove the following result concerning the structure of Pred(S).
8
Proposition 12. Fix complex numbers ν, λ, and β1 , . . . , βN and suppose S is a ν, λ-Pascal
matrix with coefficients β1 , . . . , βN . Then the following hold.
(i) The matrix Pred(S) is a banded skew-symmetric matrix with at most N nonzero bands.
(ii) The entries in each band of Pred(S) are constant, except possibly in an N + 2 × N + 2
submatrix in the bottom-right corner.
2
(iii) Pf(Pred(S)) = ω −(n−1) Pf(S), where ω = λ + ν 2 .
4
A Recurrence for the Pfaffian of a Short Sequence
Proposition 9 suggests that if {an }∞
n=1 satisfies an N + 1-term homogeneous linear recurrence
relation with constant coefficients then Pf({an }) is closely related to the Pfaffian transform
of a certain sequence of the form x1 , x2 , . . . , xN , 0, 0, . . .. With this in mind, we next study
the effect of the Pfaffian transform on such a sequence. To do this, we return to the graphtheoretic definition of the Pfaffian, which leads us to consider a particular family of graphs
that arise when we construct perfect matchings in K2n recursively.
Consider a sequence {an }∞
n=1 . Combining Definitions 2 and 5, we see that
X
Y
X
ãn =
wt(α) =
(−1)cr(α)
aj−i .
(7)
α
αK2n
(i,j)∈α
Now note that if there is an integer r for which an = 0 if n > r, then any perfect matching
α K2n which includes an edge (i, j) for which j − i > r will have wt(α) = 0. Thus when
computing ãn we need only consider perfect matchings that do not contain any such “long”
edges. With this in mind, we introduce a graph which includes only “short” edges.
Definition 13.
Suppose r and n are positive integers and x1 , . . . , xr are indeterminates.
Then the r-claw on n vertices is the weighted graph Cnr in which vertices i and j are connected
with an edge exactly when |j − i| ≤ r. If 0 < j − i ≤ r then the weight wt(i, j) of the edge
between i and j is xj−i
Suppose {an }∞
n=1 is a sequence for which an = 0 if n > r. In view of the comments
preceding Definition 13, we have
Y
X
(−1)cr(α)
ãn =
wt(i, j).
(8)
r
αC2n
(i,j)∈α
r
Each perfect matching in C2n
contains an edge (1, 1 + k) for some k, 1 ≤ k ≤ r, and when
r
we group perfect matchings in C2n
according to the value of k, equation (8) becomes


r
Y
X
X
′
(9)
(−1)cr(α ∪(1,k+1))
aj−i  .
ãn =
ak 
k=1
α′ Gk
(i,j)∈α′
r
Here Gk is the graph constructed from C2n
by removing the vertices 1 and k + 1, along with
all of their incident edges. We can repeat this process on the inner sums, obtaining yet more
9
sums over perfect matchings in graphs obtained by removing various vertices and edges from
r
C2n
. The form of these summands suggests we should generalize (8) to graphs obtained from
the claw graph by removing certain edges and vertices.
To characterize the graphs we wish to consider, suppose we start with the r-claw and
remove the edges (1, v1 ), (2, v2 ), (3, v3 ), . . . , (b, vb ), the vertices 1, v1 , 2, v2 , . . . , b, vb , and all of
the edges incident with these vertices. Note that the smallest vertex which might remain
after this process is b + 1. Since vertices in an edge differ by at most r, the largest vertex
which could be removed is b + r. The order in which 1, v1 , 2, v2 , . . . , b, vb and their incident
edges are removed does not affect the structure of the resulting graph, so the final graph is
determined by which of the vertices b + 1, b + 2, . . . , b + r remain. We introduce notation for
these graphs in the next two definitions.
Definition 14.
Fix an integer r ≥ 1. For any s, 0 ≤ s ≤ 2r − 1, the sth state of the
r-claw is the unique sequence b0 , . . . , br of 0s and 1s such that
s=
r
X
bi 2r−i .
i=0
Roughly speaking, the sth state records which of the vertices 1, 2, . . . , r we remove from
an r-claw to obtain a certain generalized r-claw.
Definition 15.
Fix an integer r ≥ 1, let x1 , . . . , xr be indeterminates, and fix an integer
r
s with 0 ≤ s ≤ 2 − 1. Let b0 , . . . , br denote the sth state of the r-claw and let b denote
the number of ones among b0 , . . . , br . Then the sth state graph of the r-claw on 2n vertices,
r,s
which we denote by C2n
, is the graph obtained from the r-claw on 2n+b vertices by removing
vertex i and all of its incident edges whenever bi−1 = 1.
Note that since we insist that 0 ≤ s ≤ 2r − 1, we always have b0 = 0. Therefore we never
r,s
r
. We can now generalize (8).
remove 1 from C2n+b
when we construct C2n
Definition 16.
Fix n ≥ 1, r ≥ 1, and s such that 0 ≤ s ≤ 2r − 1, and let x1 , . . . , xr be
s
indeterminates. Then we define the state polynomial f2n
(x1 , . . . , xr ) by
Y
X
s
(−1)cr(α)
xj−i .
(10)
f2n
(x1 , . . . , xr ) =
r,s
αC2n
(i,j)∈α
As we show next, the state polynomials include the Pfaffian transform of any sequence
{an }∞
n=1 for which an = 0 if n > r.
Proposition 17. Fix r ≥ 1 and suppose {an }∞
n=1 is a sequence for which an = 0 when n > r.
Then for all n ≥ 1 we have
0
ãn = f2n
(a1 , . . . , ar ).
(11)
r,0
r
Proof. Note that C2n
= C2n
, so the result follows by setting xi = ai for 1 ≤ i ≤ r in (10)
and comparing the result with (7).
0
In view of (11), equation (9) expresses f2n
(x1 , . . . , xr ) as a linear combination of the state
s
r
polynomials f2n−2 (x1 , . . . , xr ) for 0 ≤ s ≤ 2 − 1. In order to obtain a similar recurrence for
s
f2n
(x1 , . . . , xr ), we introduce a weight on pairs of states.
10
Definition 18.
Fix r ≥ 1 and let x1 , . . . , xr be indeterminates. Fix i, j with 1 ≤ i, j ≤
r
2 − 1, and consider the following modification of the ith state i0 , . . . , ir .
1. Set i0 = 1.
2. Choose m such that 1 ≤ m ≤ r and im = 0, and set im = 1.
3. If the current sequence begins with a 1, remove it and append a 0 to the end of the
sequence; repeat this step until the sequence begins with a 0.
If the jth state can be constructed from ith state by this process, then we define the weight
wgt(i, j) to be (−1)c xm , where c is the number of 0’s among the entries i1 , . . . , im−1 of the
ith state. Otherwise we define wgt(i, j) = 0.
Note that if we obtain the jth state from the ith state by the process described in
r,j
Definition 18 with a given m, then we obtain the state graph C2n−2
from the state graph
r,i
C2n by removing vertices 1 and m + 1, along with all of their incident edges.
Theorem 19. Fix r ≥ 1 and s such that 0 ≤ s ≤ 2r −1, and let x1 , . . . , xr be indeterminates.
If 2n ≥ r + 3 then
s
f2n
(x1 , . . . , xr )
=
r −1
2X
j
wgt(s, j)f2n−2
(x1 , . . . , xr ).
(12)
j=0
Proof. By (10) we have
s
f2n
(x1 , . . . , xr ) =
X
(−1)cr(α)
r,s
αC2n
Y
xj−i .
(13)
(i,j)∈α
r,s
Each perfect matching in C2n
has an edge of the form (1, t + 1), where 1 ≤ t ≤ r. For each
r,s
such t, let Gt denote the graph obtained from C2n
by removing vertices 1 and t + 1, along
with all of their incident edges. Now note that if α is a perfect matching which includes
(1, t + 1), then the edges in α which contain a vertex between 1 and t + 1 either cross (1, t + 1)
or involve two vertices between 1 and t + 1. Therefore, (−1)cr(α) = (−1)ct +cr(Gt ) , where ct
r,s
is the number of vertices in C2n
between 1 and t + 1. If b0 , . . . , br is the sth state then it
follows from (13) that
X
X
Y
s
f2n
(x1 , . . . , xr ) =
(−1)ct xt
(−1)cr(Gt )
xj−i .
(14)
1≤t≤s
bt =0
αGt
(i,j)∈α
r,j
We have observed that Gt = C2n−2
, where the jth state is obtained from the sth state by
the process described in Definition 18, using m = t, so the result follows from (14).
It is rare that we can apply Theorem 19 directly to the Pfaffian transform of a sequence,
since the matrices we encounter usually have a submatrix in their lower-right corner in which
the bands parallel to the main diagonal are not constant. To handle these matrices, we will
need the following generalization of Definition 16 and Theorem 19.
11
Definition 20.
Fix r ≥ 1 and N with 0 ≤ N ≤ r + 1, let x1 , . . . , xr be indeterminates,
and let B denote a skew-symmetric N × N matrix. For all n ≥ N , let Tn denote the n × n
skew-symmetric matrix with the following entries.
1. If i ≤ n − N and 1 ≤ j − i ≤ r then (Tn )ij = xj−i .
2. If i ≤ n − N and j − i > r then (Tn )ij = xj−i .
3. If j ≥ i > n − N then (Tn )ij = Bi−n+N,j−n+N .
Then for all s with 0 ≤ s ≤ 2r − 1 we define the generalized state polynomial by
X
Y
s
fB,2n
(x1 , . . . , xr ) =
(−1)cr(α)
(T2n )ij .
r,s
αC2n
(i,j)∈α
s
s
Note that if B is the 0 × 0 empty matrix then fB,2n
(x1 , . . . , xr ) = f2n
(x1 , . . . , xr ).
Theorem 21. Fix r ≥ 1, fix N with 0 ≤ N ≤ r + 1, fix s with 0 ≤ s ≤ 2r − 1, and let
x1 , . . . , xr be indeterminates. Let B denote a skew-symmetric N ×N matrix. If 2n ≥ N +r+3
then
r −1
2X
j
s
fB,2n
(x1 , . . . , xr ) =
wgt(s, j)fB,2n−2
(x1 , . . . , xr ).
(15)
j=0
Proof. This is similar to the proof of Theorem 19.
5
The Pfaffian Recurrence Theorem
Fix r ≥ 1 and suppose {an }∞
n=1 is a sequence for which an = 0 if n > r. In this section we
interpret (15) as a matrix equation, which allows us to apply the transfer matrix method
to obtain a rational generating function for ãn . To do this, we need to construct a directed
graph whose vertices are states and in which an edge from state i to state j has weight
wgt(i, j). Before introducing this graph, we note that the only states one can obtain from
state 0 by the process in Definition 18 are the even states. Therefore we use only these states
in our definition of the state digraph.
Definition 22.
For any r ≥ 1 the state digraph Dr is the weighted directed graph on
r−1
vertices 0, 1, . . . , 2 − 1 in which there is a directed edge from i to j with weight wgt(2i, 2j)
exactly when wgt(2i, 2j) 6= 0. We write Ar to denote the weighted adjacency matrix for Dr ,
which has (Ar )ij = wgt(2i, 2j) for all i, j with 0 ≤ i, j ≤ 2r−1 − 1.
We can now rewrite (15) in terms of Ar , the adjacency matrix for the state digraph.
Proposition 23. Fix r ≥ 1, fix N with 0 ≤ N ≤ r + 1, let x1 , . . . , xr be indeterminates,
s
s
and let B denote an N × N skew-symmetric matrix. Abbreviating fB,m
= fB,m
(x1 , . . . , xr ),
for all n with 2n ≥ r + 3 + N we have

 0
 0 
fB,2n−2
fB,2n

f 2
f2 
 B,2n−2 
 B,2n 
(16)
=
A
 .. 
. .
r
 .. 
 . 
r
r
2 −2
fB,2n−2
2 −2
fB,2n
12
Proof. Note that if i is even and wgt(i, j) 6= 0 then j is also even, by Definition 18. Now the
result follows from (15) and Definition 22.
We now recall the transfer matrix method, which we paraphrase from [13].
Proposition 24. [13, Theorem 4.7.2] Suppose D is a weighted directed graph with vertices
1, 2, . . . , n and adjacency matrix A, so that Aij is the weight of the edge from i to j. Then
for all i, j with 1 ≤ i, j ≤ n we have
(−1)i+j det(I − tA; j, i)
,
(A )ij t =
det(I − tA)
n≥0
X
n
n
(17)
where det(B, j, i) is the determinant of the matrix obtained by removing the jth row and ith
column of B and I is the n × n identity matrix.
Combining Propositions 23 and 24 allows us to prove our main result, which says that the
set of sequences with rational generating functions is closed under the Pfaffian transform.
Theorem 25. Suppose {an }∞
n=1 has a rational generating function. Then Pf({an }) also has
a rational generating function.
Proof. Since the generating function for {an }∞
n=1 is rational, there exist integers k, N ≥ 1
and numbers β1 , . . . , βN such that
an =
N
X
βm an−m
m=1
(n ≥ N + k).
By Proposition 9 and Definition 20 there exists an N × N skew-symmetric matrix B and
numbers b1 , . . . , bN +k such that
0
ãn = fB,2n
(b1 , . . . , bN +k )
(n ≥ N + k).
Now it follows from (16) that
ãn =
2N +k−1
X −1
j
−k
)0j fB,2N
(An−N
N +k
+2k (b1 , . . . , bN +k )
j=0
(n ≥ N + k),
so we have
∞
X
n=N +k
n
N +k
ãn t = t
2N +k−1
X −1
j
fB,2N
+2k (b1 , . . . , bN +k )
∞
X
−k
(An−N
)0j tn−N −k .
N +k
n=N +k
j=0
By Proposition 24 each term on the right is a rational function of t, so the generating function
∞
X
ãn tn is also a rational function of t.
n=1
13
6
Examples
In this section we apply the techniques we have developed to compute Pf({an }) for a variety
of interesting sequences {an }∞
n=1 . We will find it useful to have weighted adjacency matrices
for the digraphs D1 and D2 , which are as follows.
x1 −x2
(18)
A2 =
A1 = x1
x2 0
In our first example, we compute the Pfaffian transform of {cn−1 }∞
n=1 , where c is a constant.
Example 26.
Suppose c is a constant and an = cn−1 for all n ≥ 1. Then we routinely
find that for all k ≥ 1,


0
1 ··· 0 0


...
−1 0
0 0

 . .
. . . . . . . . ... 
red(Sk ) = 
.
 ..


...
0
0
0 1
0
0 · · · −1 0
0
Therefore ãn = f2n
(1) for all n ≥ 1. Since A1 = (x1 ), it follows from (12) that ãn = 1 for all
n ≥ 1.
Example 26 allows us to find the Pfaffian transform of any constant sequence.
Example 27.
Suppose c is a constant and an = c for all n ≥ 1. By Example 26, when
c = 1 we have ãn = 1 for all n ≥ 1. Combining this with Proposition 6, we find that ãn = cn
in general.
We can also generalize Example 26 to Pascal matrices.
Fix n ≥ 1 and suppose is

0
aν + λ

−(aν + λ)
0

.
.

..
..


Pred(S) = 
0
0



0
0



0
0
0
0
Example 28.
Then
where ω = λ + ν 2 . Therefore Pf(S) = ω
a 2n × 2n ν, λ-Pascal matrix with coefficient c.

···
0
0
0
0

...
0
0
0
0
..
..
.. 
...
...
.
.
.


...
0
aν + λ
0
0
,

...
aν+λ
√
0
−(aν + λ)
0
4ω


...
aν+λ
0
− √
0
1
4
ω
···
(n−1)2
0
(aν + λ)n−1 .
0
−1
0
In Example 26 we found Pf({an }) for a family of sequences satisfying a two-term linear
homogeneous recurrence relation. A natural next step is to compute the Pfaffian transform
of the most well known sequence satisfying a three-term linear homogeneous recurrence
relation, the sequence of Fibonacci numbers. To do this, we first consider a more general
family of sequences.
14
Example 29.
Suppose a and b are constants, set a1 = 1, a2 = a, and an = aan−1 + ban−2
for all n ≥ 3. Then we routinely find that for all k ≥ 1,


0
b+1
0
···
0
0
0
−b − 1
0
b + 1 ···
0
0
0


 0

−b
−
1
0
·
·
·
0
0
0


 ..
..
..
..  .
...
...
...
red(Sk ) =  .
.
.
.


 0

0
0
·
·
·
0
b
+
1
0


 0
0
0
· · · −b − 1
0
1
0
0
0
···
0
−1 0
0 1
. Since A1 = (x1 ), it follows
Therefore ãn =
+ 1) for n ≥ 2, where B =
−1 0
from (16) that ãn = (b + 1)ãn−1 for n ≥ 2, so ãn = (b + 1)n−1 for n ≥ 1. In particular, ãn is
independent of a. Setting a = b = 1, we find that the Pfaffian transform of the sequence of
Fibonacci numbers is {2n−1 }∞
n=1 .
0
fB,2n
(b
We can also generalize Example 29 to Pascal matrices.
Example 30.
Fix n ≥ 1 and suppose is a 2n × 2n ν, λ-Pascal matrix with coefficients a, b. Then Pred(S) has two nonzero bands: the first band above the diagonal
2)
2)
√
, 1 and the second band above the
, . . . , (aν+λ+b)(λ+2ν
, aν + λ + b, aν+λ+b
is (aν+λ+b)(λ+2ν
4ω
ω
ω
diagonal is
2
ν(aν+λ+b)
√
√
√
, 0,
, . . . , ν(aν+λ+b)
, ν(aν+λ+b)
4 3
ω
ω
ω
n−1
ω (n−1) (aν + λ + b)
where ω = λ + ν 2 . Therefore Pf(S) =
.
In Example 29 the matrix red(Sk ) has one fewer nonzero band than the maximum that
Proposition 9 predicts. This occurs because the extended sequence 0, 1, a, a2 +b, a3 + 2ab, . . .,
which includes the diagonal entries of Sk , continues to satisfy the recurrence an = aan−1 +
ban−2 . In the next example we compute the Pfaffian transform of the sequence of Lucas
numbers, which satisfy the Fibonacci recurrence relation, but whose initial conditions are
not compatible with prepending a 0 to the sequence.
Example 31.
Suppose a1 = 2, a2 = 1, and an = an−1 +an−2 for n ≥ 3. Then we routinely
find that for all k ≥ 1,


0
5 −1 0 · · · 0
0
0
0
−5 0
5 −1 · · · 0
0
0
0


 1 −5 0

5
·
·
·
0
0
0
0




...
0
0
0
0
0
1 −5 0
 .
..
.. . . . . . .
..
..
.. 
.
red(Sk ) = 
.
.
.
.
.
.
.
. 
 .
.


...
0
0
5 −1 0 
0
0
0


0

0
0
0
·
·
·
−5
0
5
−1


0
0
0
0 · · · 1 −5 0
2
0
0
0
0 ··· 0
1 −2 0
15
0 2
. By Proposition 24 and the
Therefore ãn =
for n ≥ 2, where B =
−2 0
2
expression for A2 in (18), the denominator of the generating function for {ãn }∞
n=1 is 1−5t+t .
In particular, ã1 = 2, ã2 = 9, and ãn = 5ãn−1 − ãn−2 for n ≥ 3.
0
fB,2n
(5, −1)
Having found the Pfaffian transform for a variety of sequences satisfying a three-term
recurrence relation, we now turn our attention to sequences satisfying a four-term recurrence
relation. We begin with the so-called Tribonacci numbers, which satisfy the recurrence
an = an−1 + an−2 + an−3 .
Example 32.
Suppose a1 = a2 = 1, a3
Then we routinely find that for all k ≥ 1,

0
2
1
−2 0
2

−1 −2 0

 0 −1 −2
 .
.. . .

.
.
red(Sk ) =  ..

0
0
0

0
0
0

0
0
0
0
0
0
= 2, and an = an−1 + an−2 + an−3 for n ≥ 4.
0
1
2
0
...
···
···
···
···
...
...
0
0
0
0
..
.
0
0
0
0
..
.
0
0
0
0
..
.

0
0

0

0


.

0
2
1 0
0

0 · · · −2 0
2 0

0 · · · −1 −2 0 1
0 ··· 0
0 −1 0


0
2 0
0
Therefore ãn = fB,2n
(2, 1) for n ≥ 3, where B = −2 0 1 . By Proposition 24 and
0 −1 0
the expression for A2 in (18), the denominator of the generating function for {ãn }∞
n=1 is
2
1 − 2t + t . Since ã1 = 1, ã2 = 2, and ã3 = 3, it is routine to show that ãn = n for all n ≥ 1.
We conclude our examples by generalizing Example 32 in a somewhat more combinatorial
direction. To set the stage for this generalization, we first recall the matchings polynomial
of a graph.
Suppose G is a graph with n vertices, and recall that a k-matching in G is a set of k edges
in G, no two of which share a vertex. Set p(G, 0) = 1, and let p(G, k) denote the number of
k-matchings in G for all k ≥ 1. Then the matchings polynomial µ(G, x) of G is defined by
X
µ(G, x) =
(−1)k p(G, k)xn−2k .
k≥0
For instance, the matchings polynomial for the path Pn with n vertices is [8, p. 2]
X
r n−k
x2n−k .
µ(Pn , x) =
(−1)
k
k≥0
For more information on the matchings polynomial, see [8, Chap. 1]. Here we give the
Pfaffian transform of a certain family of sequences in terms of this polynomial.
16
Proposition 33. Suppose a, b, and c 6= 0 are constants and set a1 = 1, a2 = a, and
a3 = a2 + b. For all n ≥ 4, set an = aan−1 + ban−2 + can−3 . Then
n−1−2k
X
b+1
b+1
n+1
k n−1−k
n+1
ãn = c µ Pn−1 ,
(−1)
=c
k
c
c
k≥0
for all n ≥ 1.
Proof. First note that

0
b+1
c
0
···
0
0
0
−b − 1
0
b+1
c
···
0
0
0

 −c
−b
−
1
0
b
+
1
·
·
·
0
0
0


...
 0
−c
−b − 1
0
0
0
0
 .
.
.
.
..
.
.
.
..
..
..
..
..
..
red(Sn ) = 
.
 ..

.
..
 0
0
b+1
c
0
0
0

 0
0
0
0
·
·
·
−b
−
1
0
b
+
1

 0
0
0
0
···
−c
−b − 1
0
0
0
0
0
···
0
0
−1
√
Now let c be any square root of c and observe that
√
√
red(Sn ) = ( cI)Tn ( cI)t ,
where

0
b+1
c
1
b+1
− b+1
0
c
 c
 −1 − b+1
0
c


 0
−1 − b+1
c
 .
..
..
.
Tn = 
.
.
 .

 0
0
0

 0
0
0

 0
0
0
0
0
0
0
1
b+1
c
0
...
0
0
0
0
···
···
···
...
...
0
0
0
0
0
0
0
...
0
..
.
...
0
b+1
c
···
···
···
− b+1
c
−1
0
0
− b+1
c
0

0
0

0


0
.. 
.
.

0

0

1
0
(19)

0
0

0


0 0
.. 
..
.
.
.

1 0

b+1
0

c
0 c
−c 0
0
0
0
To compute Pf(Tn ), first notice that any perfect matching in K2n which contributes to Pf(Tn )
must include the edge (2n−1, 2n), which will cross no other edges in the matching. Therefore
b+1
0
Pf(Tn ) = cf2n−2
,1 .
(20)
c
2
We now see that Pf(Tn ) is a sum over perfect matchings in C2n−2
; to each such perfect
matching we associate a matching (not necessarily perfect) in the path Pn−1 with vertices
2
v1 , . . . , vn−1 . In particular, given a perfect matching α C2n−2
, the associated matching α′
in Pn−1 consists of those edges (vi , vi+1 ) for which (2i − 1, 2i + 1) is an edge in α. Since α is
2
a matching in C2n−2
, no two edges in α′ share a vertex, so α′ is a matching in Pn−1 .
17
To show that we have a bijection between matchings in Pn−1 and perfect matchings in
suppose α′ is a matching in Pn−1 . To construct α, first construct edges (2i − 1, 2i + 1)
for each edge (vi , vi+1 ) in α′ . Now suppose 2i − 1 is not yet incident with any edges in our
partial α. If (2i − 2, 2i − 1) were an edge in α, then α would contain a perfect matching
2
2
in C2i−3
. But this is impossible, since C2i−3
has an odd number of vertices. Therefore to
construct α we must connect every as-yet unmatched odd vertex 2i − 1 with the even vertex
2i. Now suppose some even vertex 2i is not yet incident with any edges in our partial α.
Then we must already have added (2i − 1, 2i + 1) or (2i − 3, 2i − 1) to α. In the first case,
adding (2i − 2, 2i) to α forces a perfect matching of a graph with an odd number of vertices.
However, 2i + 1 is not connected with 2i + 2, so we must include (2i, 2i + 2) in α. By similar
reasoning, in the second case we must include (2i − 2, 2i) in α. At this point every vertex
2
in C2n−2
is incident with exactly one edge in α, so α is a perfect matching. Moreover, by
construction α′ is the perfect matching in Pn−1 associated with α.
2
To complete the proof, suppose α is a perfect matching in C2n−2
and α′ is the associated
matching in Pn−1 . If α′ has k edges then α has exactly k edges of the form (2i − 1, 2i + 1),
exactly k edges of the form (2i, 2i + 2), and n − 1 − 2k edges of the form (2i − 1, 2i). By
construction each edge of the form (2i − 1, 2i + 1) yields a crossing in α, and all crossings
in α are of this type. Moreover, the only edges in α whose weight is not 1 are those of the
form (2i − 1, 2i), each of which has weight b+1
. Combining these observations, we find that
c
2
C2n−2
,
0
f2n−2
n−1−2k
X
b+1
b+1
k
.
,1 =
(−1) p(Pn−1 , k)
c
c
k≥0
(21)
Putting everything together, we have
as desired.
7
ãn = Pf(red(Sn ))
√
√
= Pf(( cI)Tn ( cT )t )
√
= det( cI) Pf(Tn )
b+1
0
,
1
= cn+1 f2n−2
c
= cn+1 µ Pn−1 , b+1
c
(by Prop. 9)
(by (19))
(by Prop. 3)
(by (20))
(by (21)),
Acknowledgment
The authors thank the anonymous referee who described how their results could be generalized to compute the Pfaffian of a Pascal matrix.
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2000 Mathematics Subject Classification: Primary 15A15; Secondary 05C50, 05C70.
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Keywords: Pfaffian, skew-symmetric matrix, Toeplitz matrix, transfer matrix method, perfect matching, Hankel transform.
(Concerned with sequences A000045, A000073, A000079, A000108, A000244, A001006, and
A001045.)
Received June 8 2008; revised version received December 16 2008. Published in Journal of
Integer Sequences, December 17 2008.
Return to Journal of Integer Sequences home page.
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